Problem: Jeff will pick a card at random from ten cards numbered 1 through 10.  The number on this card will indicate his starting point on the number line shown below.  He will then spin the fair spinner shown below (which has three congruent sectors) and follow the instruction indicated by his spin.  From this new point he will spin the spinner again and follow the resulting instruction.  What is the probability that he ends up at a multiple of 3 on the number line?  Express your answer as a common fraction. [asy]
import graph;
size(10cm);
defaultpen(linewidth(0.7)+fontsize(8));

xaxis(-2,13,Ticks(OmitFormat(-1),1.0,begin=false,end=false,beginlabel=false,endlabel=false),Arrows(4));

label("-1",(-1,-0.98));

real r=3.5;
pair center=(17,0);
draw(circle(center,r));
int i;
for(i=1;i<=3;++i)

{

draw(center--center+r*dir(120*i-30));

}
label("$\parbox{1cm}{move \\ 1 space \\ left}$",center+r/2*dir(150));
label("$\parbox{1cm}{move \\ 1 space \\ right}$",center+r/2*dir(270));
label("$\parbox{1cm}{move \\ 1 space \\ right}$",center+r/2*dir(30));
draw(center--center+3*r/4*dir(80),EndArrow(4));[/asy]
Answer: Use two-letter strings to denote the results of the two spins.  For example, RL denotes spinning ``move one space right'' followed by ``move one space left.'' If Jeff starts at a multiple of 3, the only ways he can end up at a multiple of 3 are to spin LR or RL.  The probability of starting at a multiple of 3 is $\frac{3}{10}$, and the probability of spinning LR or RL is $\frac{1}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{9}$.  Multiplying these probabilities, we find that the probability that Jeff will start at a multiple of 3 and reach a multiple of 3 is $\frac{12}{90}$.

If Jeff starts at a number which is one more than a multiple of 3, the only way for him to reach a multiple of 3 for him to spin RR.  The probability of selecting 1, 4, 7, or 10 is $\frac{4}{10}$, and the probability of spinning RR is $\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$.  The probability that Jeff will start one unit to the right of a multiple of 3 and end up at a multiple of 3 is $\frac{16}{90}$.

If Jeff starts at a number which is one less than a multiple of 3, the only way for him to reach a multiple of 3 for him to spin LL.  The probability of selecting 2, 5, or 8 is $\frac{3}{10}$, and the probability of spinning LL is $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$.  The probability that Jeff will start one unit to the left of a multiple of 3 and end up at a multiple of 3 is $\frac{3}{90}$.

In total, the probability that Jeff will reach a multiple of 3 is $\dfrac{12}{90}+\dfrac{3}{90}+\dfrac{16}{90}=\boxed{\frac{31}{90}}$.